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(D) Let $(x, y)$ be the coordinates in the original system and $(x', y')$ be the coordinates in the new system after rotation by an angle $	heta = 13

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$\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$

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(D) Let $(x, y)$ be the coordinates in the original system and $(x', y')$ be the coordinates in the new system after rotation by an angle $	heta = 135^{\circ}$.The transformation equations are:$x' = x \cos 	heta + y \sin 	heta$$y' = -x \sin 	heta + y \cos 	heta$Given $(x', y') = (4, -3)$ and $	heta = 135^{\circ}$,we have $\cos 135^{\circ} = -\frac{1}{\sqrt{2}}$ and $\sin 135^{\circ} = \frac{1}{\sqrt{2}}$.Substituting these values:$4 = x(-\frac{1}{\sqrt{2}}) + y(\frac{1}{\sqrt{2}}) \implies -x + y = 4\sqrt{2} \quad (i)$$-3 = -x(\frac{1}{\sqrt{2}}) + y(-\frac{1}{\sqrt{2}}) \implies -x - y = -3\sqrt{2} \implies x + y = 3\sqrt{2} \quad (ii)$Adding $(i)$ and $(ii)$:$2y = 7\sqrt{2} \implies y = \frac{7\sqrt{2}}{2} = \frac{7}{\sqrt{2}}$Subtracting $(i)$ from $(ii)$:$2x = -\sqrt{2} \implies x = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$Thus,the original coordinates are $(x, y) = \left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}ight)$.

The coordinate axes are rotated through an angle $135^{\circ}$. If the coordinates of a point $P$ in the new system are known to be $(4, -3)$,then the coordinates of $P$ in the original system are

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